Chemistry, 29.07.2020 01:01 aesthvx44

Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer coefficients. The reaction that takes place when chlorine gas combines with aqueous potassium bromide. (Use the lowest possible coefficients. Omit states of matter.)

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Answer from: dondre54

$\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl$ .

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

Formula for each of the species

Start by finding the formula for each of the compound.

Both chlorine $\rm Cl$ and bromine $\rm Br$ are group 17 elements (halogens.) Each On the other hand, potassium $\rm K$ is a group 1 element (alkaline metal.) Each

Therefore, the ratio between $\rm K$ atoms and $\rm Br$ atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula $\rm KBr$ . Similarly, the ratio between

The formula for chlorine gas is $\rm Cl_2$ , while the formula for bromine gas is $\rm Br_2$ .

Balanced equation for the reaction

Write down the equation using these chemical formulas.

$\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl$ .

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both $\rm KBr$ and $\rm KCl$ features two elements each.

Assume that the coefficient of $\rm KCl$ is one. Hence:

$\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl$ .

Note that $\rm KBr$ is the only source of $\rm K$ and $\rm Br$ atoms among the reactants of this reaction.

There would thus be one $\rm K$ atom and one $\rm Br$ atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of $\rm K$ and $\rm Br$ atoms on the product side of the equation.

In this reaction, $\rm Br_2$ is the only product with $\rm Br$ atoms.

One $\rm Br$ atom would correspond to 0.5 units of $\rm Br_2$ .

Similarly, in this reaction, $\rm KCl$ is the only product with $\rm K$ atoms.

One $\rm K$ atom would correspond to one formula unit of $\rm KCl$ .

Hence:

$\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl$ .

Similarly, there should be exactly one $\rm Cl$ atom on either side of this equation. The coefficient of $\rm Cl_2$ should thus be 0.5 . Hence:

$\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl$ .

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

$\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl$ .

Answer from: Quest

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

Answer from: Quest

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante

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